Prof. D. R. Wilton

Notes 15 Notes 15 Plane WavesPlane Waves

ECE 3317

[Chapter 3]

Introduction to Plane Waves

A plane wave is the simplest solution to Maxwell’s equations for a wave that travels through free space.

The wave does not requires any conductors – it exists in free space.

A plane wave is a good model for radiation from an antenna, if we are far enough away from the antenna.

x

z

E

H

S

S E H

http://www.impression5.org/solarenergy/misc/emspectrum.html

The Electromagnetic Spectrum

1 pm

1 fm

1 kHz

1 MHz

1 GHz

1 THz

1 PHz

1 mm

Source Frequency Wavelength

U.S. AC Power  60 Hz  5000 km

ELF Submarine Communications 500  Hz  600 km

AM radio (KTRH) 740  kHz 405 m 

TV ch. 2 (VHF)  60 MHz 5 m 

FM radio (Sunny 99.1) 99.1  MHz 3 m 

TV ch. 8 (VHF) 180  MHz 1.7 m 

TV ch. 39 (UHF) 620  MHz 48 cm 

Cell phone (PCS) 850  MHz 35 cm 

Cell Phone (PCS 1900) 1.95 GHz 15 cm

μ-wave oven 2.45 GHz 12 cm 

Police radar (X-band) 10.5 GHz 2.85 cm 

mm wave 100 GHz 3 mm 

Light 51014 [Hz] 0.60 m

X-ray 1018 [Hz] 3  Å

The Electromagnetic Spectrum (cont.)

TV and Radio Spectrum

VHF TV: 55-216 MHz (channels 2-13)

Band I : 55-88 MHz (channels 2-6)Band III: 175-216 MHz (channels 7-13)

FM Radio: (Band II) 88-108 MHz

UHF TV: 470-806 MHz (channels 14-69)

AM Radio: 520-1610 kHz

Note: Digital TV broadcast takes place primarily in UHF and VHF Band III.

Vector Wave Equation

Start with Maxwell’s equations in the phasor domain:

E H

H J E

j

j

Faraday’s law

Ampere’s law

Assume free space:

J E = 0

0

0

E H

H E

j

j

0

0

We then have

Ohm’s law:

Vector Wave Equation (cont.)

Take the curl of the first equation and then substitute the second into the first:

0

0 0

E H

E

j

j j

0 0 0k

20E E 0k Vector Wave Equation

Define

Hence

0

0

E H

H E

j

j

wavenumber of free space [rad/m]

Vector Helmholtz Equation

20 E )E (0k t

Recall the vector Laplacian identity:

2V V V

Hence we have

2 20E E E 0k

Also, from the divergence of the vector wave equation, we have:

20E E 0k

20E 0 0k if

Solutions ofautomatically satisfy Gauss’s law

( )t

Hence we have:

vector Helmholtz equation2 2

0E E 0 ( )k t t

Recall the property of the vector Laplacian in rectangular coordinates:

2 2 2 2x y zV x V y V z V

Taking the x component of the vector Helmholtz equation, we have

2 20E E 0x xk

Vector Helmholtz Equation (cont.)

scalar Helmholtz equation

( ),

( ,)

t t

t

Since we'll solve

not we'll have to

check that our solution

is divergence

less!

Plane Wave Field

Assume

Then

or

E E ( )xx z

2 20

2 2 2202 2 2

2202

E E 0

E E EE 0

EE 0

x x

x x xx

xx

k

kx y z

dk

dz

Solution: 00E ( ) jk z

x z E e 0 0 0( )k

y

z

xE

The electric field is polarized in the x direction, and the wave is propagating (traveling) in the z direction.

00E ( ) jk z

x z E e

For wave traveling in the negative z direction:

Suppose we had assumed

Then

or

ˆE E ( )zz z

2 20

2 2 2202 2 2

2202

E E 0

E E EE 0

EE 0

z z

z z zz

zz

k

kx y z

dk

dz

Solution:0

0E ( ) jk zz z E e

y

z

xE

Hence the electric field would be both polarized and vary along the z direction.

00 0 0ˆE ( ) 0 0jk z

zz z jk E e E But

• Plane waves are polarized transverse to the propagation direction• Longitudinally polarized plane waves are not possible.

Plane Wave Field (cont.)

where

0E ( ) jkzx z E e

k

For a wave traveling in a lossless dielectric medium:

0

0

r

r

wavenumber of dielectric medium

Plane Wave Field (cont.)

or 0ˆE( ) jkzz x E e

0E ( ) jkzx z E e

The electric field behaves exactly as does the voltage on a transmission line.

k Notational change:

Note: for a lossless transmission line, we have: LC

A transmission line filled with a dielectric material has the same wavenumber as does a plane wave traveling through the same material.

Plane Wave Field (cont.)

A wave travels with the same velocity on a transmission line as it does in space, provided the material is the same.

The H field is found from:

so

E Hj

1H E

E1

1( ) E

x

x

x

x zj

dy

j dz

jk yj

ˆ ˆ ˆ

E =

E E Ex y z

x y z

x y z

0E ( ) jkzx z E e

Plane Wave Field (cont.)

Intrinsic Impedance

or

Hence

H E x

ky

E EE H , H

Hx x

x y yy k

where

[ ]

Intrinsic impedance

of the medium

H Ey x

k

00

0

376.730313 [ ] 120 [ ]

Free-space:

20

L C

Z L C

GF, GF,

GF

Recall for TX lines,

( Geometrical Fac

GF

GF tor)

Poynting Vector

The complex Poynting vector is given by:

0

0

ˆE

1ˆH

jkz

jkz

x E e

y E e

*1S E H

2

Hence we have

*

00

*0 0

2

0

1ˆS

2

1ˆ

2

1ˆ

2

jkz jkz

jkz jkz

Ez E e e

z E e E e

z E

2

0 2ˆS [VA/m ]2

Ez

2

0 2ˆ= Re S [W/m ]2

ES t z

Phase Velocity

0

0 0

E

cos

jkzx

x

E e

E E t kz

pvk

1p dv c

From our previous discussion on phase velocity for transmission lines, we know that

Hence we have

(speed of light in the dielectric material)

pv

so

00 0

jE E e

Note: All plane waves travel at the same speed in a lossless medium, regardless of the frequency. This implies that there is no dispersion, which in turn implies that there is no signal distortion.

Wavelength

2 2

k

2 2 2 1

2dc

k ff f

dc

f

From our previous discussion on wavelength for transmission lines, we know that

Hence

Also, we can write

00

2

k

Free space:

Free space: 0

c

f

0

0 0

E

cos

jkzx

x

E e

E E t kz

00 0

jE E e

Summary (Lossless Case)

0

0

2

0

E

1H

S2

jkzx

jkzy

z

E e

E e

E

y

z

x E H

S

0 0cosxE E t kz

00 0

jE E e

0

0Re E Re jj t jkz j tx xE e E e e e

Time domain:

Denote

1pv

k

2

k

k

00cosy

EH t kz

Lossy Medium

E H

H J E

j

j

J E

Return to Maxwell’s equations:

Assume Ohm’s law:

H E E

E

j

j

Ampere’s law

We define an effective (complex) permittivity c that accounts for conductivity:

cj j c j

z

xE

ocean

Lossy Medium

E H

H Ec

j

j

Maxwell’s equations then become:

The form is exactly the same as we had for the lossless case, with

c

Hence we have

0

0

E

1H

jkzx

jkzy

E e

E e

' "ck k jk

j

c

e

(complex)

(complex)

Lossy Medium (cont.)

Examine the wavenumber:

' "k k jk ' "

0 0jkz jk z k z

xE E e E e e

c k

ck c j

' 0

0

k

k

Denote:

In order to ensure decay, the wavenumber k must be in the fourth quadrant.

Hence: 'k

k

Compare with lossy TL:

Note! " Im 0k k

Note! 00x xE E

Lossy Medium (cont.)

' "0

jk z k zxE E e e

z0

k zE e

2

k

Lossy Medium (cont.)

' "0

jk z k zxE E e e

1/ "pd k

The “depth of penetration” dp is defined.

"k ze

z

0/xE E 1

1 0.37e

pd

Lossy Medium (cont.)

2 2

0 0* 2 " 2 "*

1S E H

2 2 2k z j k z

x y

E Ez z e z e e

' "0

' "0

1

jk z k zx

jk z k zy

E E e e

H E e e

The complex Poynting vector is:

j

c

e

2

0 2 "ReS cos2

k zz z

ES t e

2 "k ze

z

Sz1

1/ "pd k

2 0.14e

Example

Ocean water:

0

81

4 [S/m]r

Assume f = 2.0 GHz

00

c rj j

0 81 35.95 [F/m]c j

0 0 0c rck 1386.022 81.816 [m ]k j

1/pd k 0.013 [m]pd

2 / k 0.016 [m]

(These values are fairly constant up through microwave frequencies.)

Example (cont.)

f dp [m]

1 [Hz] 251.610 [Hz] 79.6100 [Hz] 25.2 1 [kHz] 7.9610 [kHz] 2.52100 [kHz] 0.796 1 [MHz] 0.26210 [MHz] 0.080100 [MHz] 0.02621.0 [GHz] 0.01310.0 [GHz] 0.012100 [GHz] 0.012

The depth of penetration into the ocean water is shown for various frequencies.

Loss Tangent

c j

Denote: c c cj

The loss tangent is defined as: tan c

c

We then have: tan

The loss tangent characterizes the nature of the material:

tan << 1: low-loss medium (attenuation is small over a wavelength)tan >> 1: high-loss medium (attenuation is large over a wavelength)

2 / 2 /k k k kke e e Note:

c

c

jc c e

0 0

0

0

1

1 tan

ck j

j

j

Low-Loss Limit

0 1 tan / 2k j Low-loss limit: 1

21 1 , 12

( )zz z Using

The wavenumber may be written in terms of the loss tangent as

0

11

2k j

k 0

1

2

0

2k

0

2pd

(independent of frequency)

or

Polarization Loss

The permittivity can also be complex, due to molecular and atomic polarization loss (friction at the molecular and atomic levels) .

c j

is due to the motion of charges

is due to the motion of charges

bound

free

j

c j j

c j

Example: distilled water: 0 (but heats up well in a microwave oven!).

tan

or

Note: In practice, it is usually difficult to distinguish how much of the loss tangent comes from conductivity and how much comes from polarization loss.

Polarization Loss (cont.)

c j

1 tanc j

0 r

Regardless of where the loss comes from, (conductivity or polarization loss), we can write

where

1c j

or

Note: If there is no polarization loss, then r r

Polarization Loss (cont.)

For modeling purposes, we can lump all of the losses into an effective conductivity term:

effc j

c j

0 r where